# Maclaurin's Series question

• Sep 20th 2006, 01:35 AM
margaritas
Maclaurin's Series question
I can't get this:

y = ln(1 + tan x)

(i) Show that d2y/dx2 = -1 when x = 0.

Thanks if you could help, I'm preparing for my promos next week!
• Sep 20th 2006, 03:43 AM
CaptainBlack
Quote:

Originally Posted by margaritas
I can't get this:

y = ln(1 + tan x)

Step 1 use the chain rule:

dy/dx=[1/(1+tan(x))](sec(x))^2

Step 2 use product rule (you could use the quotient ruleif you prefer)
and chain rule:

d^2y/dx^2=[1/(1+tan(x))^2](-1) [sec(x)]^4 + [1/(1+tan(x))].2.[sec(x)]^2 tan(x)

Quote:

(i) Show that d2y/dx2 = -1 when x = 0.
when x=0 tan(x)=0, and sec(x)=1, result follows.

RonL
• Sep 20th 2006, 03:50 AM
Soroban
Hello, margaritas!

It's just two applications of the Quotient Rule . . .
. . Exactly where is your difficulty?

Quote:

y = ln(1 + tan x)

(i) Show that d²y/dx² = -1 when x = 0

dy . . . . . .sec²x
--- . = . ------------
dx . . . . 1 + tan x

d²y . . . (1 + tan x)·2·sec x(sec x·tan x) - sec²x·sec²x
---- .= .------------------------------------------------------
dx² . . . . . . . . . . . . . (1 + tan x)²

. . . . . . 2·sec²x·tan x(1 + tan x) - sec²x·sec²x
. . . = . ----------------------------------------------
. . . . . . . . . . . . . (1 + tan x)²

When x = 0, we have:

. . . d²y . . . . 2·sec²0·tan 0(1 + tan 0) - sec²0·sec²0
. . . ---- . = . ----------------------------------------------- . = . -1
. . . dx² . . . . . . . . . . . (1 + tan 0)²

• Sep 20th 2006, 04:12 AM
margaritas
Oh right I got it! I was probably put off by the fact that it was so tedious to convert sec^2 x to 1/cos^2x and then converting further again using the double angle formula, but anyway thanks guys for your help! :)